Suppose that I have a complex manifold $X$ with a symplectic form $\omega$ and a continuous function $f:X\to\Bbb R$ such that $\omega=2i\partial\bar{\partial}f$. Does that imply that $X$ is Kähler? If not, is there a condition on $f$ that implies $X$ is Kähler?
We can always define a two-form $g$ by $g(X,Y)=\omega(X,IY)$ but is it a Riemannian metric?
You just need $i\partial\bar\partial f>0$, i.e. $f$ is plurisubharmonic. Intuitively, $i\partial\bar\partial$ is a complex Hessian, so plurisubharmonicity is a complex convexity condition.